Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation cos(x)=√3/2
- Equation x^2-x-56=0
-
Equation x^3=-8
-
Equation 4x^2-12x+9=0
- Express {x} in terms of y where:
- 16*x-15*y=5
- 17*x-1*y=-15
- -8*x+12*y=-6
- -10*x-10*y=11
- Derivative of:
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x^3
- Integral of d{x}:
-
x^3
-
-8
- x^3
- Sum of series:
-
-8
- Limit of the function:
- -8
-
Identical expressions
- x^ three =- eight
- x cubed equally minus 8
- x to the power of three equally minus eight
- x3=-8
- x³=-8
- x to the power of 3=-8
-
Similar expressions
- x^3=+8
x^3=-8 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$x^{3} = -8$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{x^{3}} = \sqrt[3]{-8}$$
or
$$x = 2 \sqrt[3]{-1}$$
Expand brackets in the right part
We get the answer: x = 2*(-1)^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = -8$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -8$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = -1$$
so
$$\cos{\left(3 p \right)} = -1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = -2$$
$$z_{2} = 1 - \sqrt{3} i$$
$$z_{3} = 1 + \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = -2$$
$$x_{2} = 1 - \sqrt{3} i$$
$$x_{3} = 1 + \sqrt{3} i$$
$$x^{3} = -8$$
Because equation degree is equal to = 3 - does not contain even numbers in the numerator, then
the equation has single real root.
Get the root 3-th degree of the equation sides:
We get:
$$\sqrt[3]{x^{3}} = \sqrt[3]{-8}$$
or
$$x = 2 \sqrt[3]{-1}$$
Expand brackets in the right part
x = -2*1^1/3
We get the answer: x = 2*(-1)^(1/3)
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{3} = -8$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{3} e^{3 i p} = -8$$
where
$$r = 2$$
- the magnitude of the complex number
Substitute r:
$$e^{3 i p} = -1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(3 p \right)} + \cos{\left(3 p \right)} = -1$$
so
$$\cos{\left(3 p \right)} = -1$$
and
$$\sin{\left(3 p \right)} = 0$$
then
$$p = \frac{2 \pi N}{3} + \frac{\pi}{3}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = -2$$
$$z_{2} = 1 - \sqrt{3} i$$
$$z_{3} = 1 + \sqrt{3} i$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = -2$$
$$x_{2} = 1 - \sqrt{3} i$$
$$x_{3} = 1 + \sqrt{3} i$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 8$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = 8$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 8$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = 8$$
Sum and product of roots
[src]
sum
___ ___ -2 + 1 - I*\/ 3 + 1 + I*\/ 3
$$\left(-2 + \left(1 - \sqrt{3} i\right)\right) + \left(1 + \sqrt{3} i\right)$$
=
0
$$0$$
product
/ ___\ / ___\ -2*\1 - I*\/ 3 /*\1 + I*\/ 3 /
$$- 2 \left(1 - \sqrt{3} i\right) \left(1 + \sqrt{3} i\right)$$
=
-8
$$-8$$
-8
Rapid solution
[src]
x1 = -2
$$x_{1} = -2$$
___ x2 = 1 - I*\/ 3
$$x_{2} = 1 - \sqrt{3} i$$
___ x3 = 1 + I*\/ 3
$$x_{3} = 1 + \sqrt{3} i$$
x3 = 1 + sqrt(3)*i
Numerical answer
[src]
x1 = 1.0 - 1.73205080756888*i
x2 = -2.0
x3 = 1.0 + 1.73205080756888*i
x3 = 1.0 + 1.73205080756888*i
The graph